The Hilbert transform along a one variable vector field Christoph Thiele (joint work with M. Bateman) Conference in honor of Eli Stein, Princeton, 2011

Partial list of work by Eli Stein that had impact on this research -Stein: Problems in harmonic analysis related to curvature and oscillatory integrals, Proc ICM Phong, Stein: Hilbert integrals, singular integrals, and Radon transforms II, Invent. Math, Christ, Nagel, Stein, Wainger: Singular and maximal Radon transforms. Annals of Math, Stein, Wainger: Oscillatory integrals related to Carleson’s theorem, Math Research Lett, Stein, Street: Multi-parameter singular Radon transforms, preprint, 2011

Outline of lecture 1)Hilbert transform along vector fields with a) regularity (analytic, Lipshitz) condition b) one variable condition (main topic here) 2)Connection with Carleson’s theorem 3)Reduction to covering lemmas 4)Three different covering lemmas

Vector Field in the Plane

Hilbert Transform/Maximal Operator along Vector Field Question: - bounds

First observations Bounded by 1D result if vector field constant. Value independent of length of v(x,y). May assume unit length vector field. Alternatively, may assume v(x,y)=(1,u(x,y)) for scalar slope field u.

Nikodym set example Set E of null measure containing for each (x,y) a line punctured at (x,y). If vector field points in direction of this line then averages of characteristic fct of set along vf are one.

Gravitation vector field HT/MO of bump function asymptotically, only for p>2, weak type 2

Propose modification/conditions Truncate integral at (normalize ) Demand slow rotation (Lipshitz: )

Zygmund/Stein conjectures Assume,, and truncate. Conj.1:Truncated MO bounded. Conj.2:Truncated HT bounded. No bounds known except 1) if p is infinity.

Analytic vector fields If v is real analytic, then on a bounded domain: Bourgain (1989): MO is bounded in, p>1. Christ, Nagel, Stein, Wainger (1999): HT bounded in (assume no straight integral curves. Stein,Street announce without assumption)

One variable (meas.) vector field

Theorem (M. Bateman, C.T.) Measurable, one variable vector field (HT not truncated; Related earlier work: Bateman; Lacey/Li)

Linear symmetry group Isotropic dilations Dilation of second variable Shearing

Constant along Lipschitz Angle of to x axis less than Angle of to x-axis less than or equal to Conjecture: Same bounds for as in Bateman,CT

Relation with Carleson’s theorem and time-frequency analysis

Carleson’s operator Carleson 1966, Hunt 1968: Carleson’s operator is bounded in,

Coifman’s argument

The argument visualized

A Littlewood Paley band

Bound for supported on Littlewood Paley band Lacey and Li: Bound on for arbitrary two variable measurable vector field Bateman: Bound on for one variable vector field.

Vector valued inequality, reduction to covering arguments

Littlewood Paley decomp.

Vector valued inequality Since LP projection commutes with HT (vector field constant in vertical direction), Enough to prove for any sequence

Weak type interpolation Enough to prove for Whenever

Cauchy Schwarz Enough to prove Which follows from

Single band operator estimate By interpolation enough for Lacey-Li (p>2) /Bateman (p<2) proved Note: F,E depend on k, while G,H do not

Induction on. If p<2 and Find such that and the desired estimate holds (prove!) for. Apply induction hypothesis on (gain )

Induction on. If p>2 and Find such that and And the desired estimate holds for Apply induction hypothesis on

Finding exceptional sets. Covering arguments

Parallelograms

Kakeya example Try union of all parallelograms R with for appropriate. Bad control on, example of Kakeya set.

Vector field comes to aid Restrict attention to U( R), set of points in R where the direction of the vector field is in angular interval E( R) of uncertainty of R

1st covering lemma The union of all parallelograms with has measure bounded by for q>1 (vector field measurable, no other assumption)

Outline of argument Find good subset of set of parallelograms with large density, such that Then:

Greedy selection Iteratively select R for with maximal shadow such that for previously selected R’ Here 7R means stretch in vertical direction.

Vertical maximal function The non-selected parallelograms are all inside which has acceptable size.

Additional property R’ selected prior to R; U(R’) intersect U(R). Then

argument Assume in order of selection and

2nd covering lemma (Lacey-Li) The union of all parallelograms with and Has measure bounded by. Use vector field Lipshitz in vertical direction. -power responsible for.

Outline of argument From set of parallelograms with large densities select as before. Using as before obtain: To prove: Power of 2 here responsible for -power

Expansion of. Sum over all pairs with R selected before R’. Case 1 Case 2 Second case has aligned directions, as before.

Second selection in Case 1 Fix R, consider R’ selected later with Case 1. Prove Select so that each selected R’ has Projection of U(R’) onto x-axis disjoint from Projection of U(R’’) for previously selected R’’

Removing δ Use disjointness of projections of U(R’) and density δ to reduce to showing for fixed R’ Summing over those R’’ in which where not selected for because of prior selection of R’. All R’’ have similar angle as R’. (use vector field Lipschitz in vertical direction.)

Picture of situation

Back to maximal function 1) Hard case is when all rectangles are thin. 2) Intersection with R is only fraction α (depending on angle) of R 3) If too much overlap, then vertical maximal function becomes too large in extended rectangle (1/α) R. 4) Two effects of α cancel.

3rd covering lemma (Bateman) One variable vf, parallelograms of fixed height h. Union of all parallelograms R with has measure bounded by

Difference to previous situation Single height and constant vf in vertical direction causes approximately constant slope for all R’’ and thus avoids overlap.